Spectral calculus, Gaussian fields and covariance normalization prerequisites

For $u\in D(A)$, [F5] and nonnegativity give $\mathrm{\parallel }(I+A)u\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }\ge \mathrm{\mid }⟨(I+A)u,u⟩\mathrm{\mid }=\mathrm{\parallel }u{\mathrm{\parallel }}^2+⟨Au,u⟩\ge \mathrm{\parallel }u{\mathrm{\parallel }}^2$. Hence $I+A$ is injective and bounded below by $1$. To prove $A$ is closed, suppose $u_n\in D(A)$, $u_n\to u$, and $A{u}_n\to v$. For every $w\in D(A)$, self-adjointness gives $⟨Aw,u⟩={\lim }_n⟨Aw,u_n⟩={\lim }_n⟨w,A{u}_n⟩=⟨w,v⟩$. In particular, $\mathrm{\mid }⟨Aw,u⟩\mathrm{\mid }=\mathrm{\mid }⟨w,v⟩\mathrm{\mid }\le \mathrm{\parallel }w\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ by [F5], so this functional is bounded in the ambient $H$ norm on $D(A)$. By [F6], $u\in D(A^{\ast })$ and its unique representing vector is $A^{\ast }u=v$. Self-adjointness, including equality of domains, yields $u\in D(A)$ and $Au=v$. Thus $A$ is closed. For $w\in D((I+A{)}^{\ast })$, write $z=(I+A{)}^{\ast }w$. Subtracting the identity term gives $⟨Au,w⟩=⟨u,z-w⟩$ for every $u\in D(A)$. By [F5] this is bounded by $\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }z-w\mathrm{\parallel }$, so [F6] gives $w\in D(A^{\ast })$ and $A^{\ast }w=z-w$, or $(I+A{)}^{\ast }w=w+A^{\ast }w$. Conversely, if $w\in D(A^{\ast })$, then $⟨(I+A)u,w⟩=⟨u,w+A^{\ast }w⟩$ for every $u\in D(A)$. The functional in $u$ is bounded by $\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }w+A^{\ast }w\mathrm{\parallel }$ by [F5], so [F6] gives $w\in D((I+A{)}^{\ast })$ with $(I+A{)}^{\ast }w=w+A^{\ast }w$. Thus $(I+A{)}^{\ast }=I+A^{\ast }=I+A$ with domain $D(A)$, so $I+A$ is self-adjoint. If $u_n\in D(A)$ and $(I+A)u_n$ converges, the lower bound makes $(u_n)$ Cauchy; closedness then puts the limit in the range, so the range of $I+A$ is closed. If $y$ is orthogonal to that range, then $⟨(I+A)x,y⟩=0=⟨x,0⟩$ for all $x\in D(A)$; [F6] with representing vector zero gives $y\in \ker (I+A{)}^{\ast }=\ker (I+A)$, which is zero by the same lower bound; the range is therefore dense and hence all of $H$. Thus $B=(I+A{)}^{-1}$ exists everywhere, is bounded with norm at most $1$, is injective, and has range $D(A)$, which is dense.

If $x=(I+A)u$ and $y=(I+A)v$, then $⟨Bx,y⟩=⟨u,(I+A)v⟩=⟨(I+A)u,v⟩=⟨x,By⟩$, so $B$ is self-adjoint. Also $⟨Bx,x⟩=⟨u,(I+A)u⟩=\mathrm{\parallel }u{\mathrm{\parallel }}^2+⟨Au,u⟩\ge 0$, so $B$ is positive. For nonreal $z$, Cauchy–Schwarz [F5] and self-adjointness give $\mathrm{\parallel }(B-zI)w\mathrm{\parallel }\ge \mathrm{\mid }\mathrm{Im}z\mathrm{\mid }\mathrm{\parallel }w\mathrm{\parallel }$ and the same lower bound for its adjoint, so its range is closed and dense. For real $z<0$, positivity and [F5] give $\mathrm{\parallel }(B-zI)w\mathrm{\parallel }\ge (-z)\mathrm{\parallel }w\mathrm{\parallel }$, again giving closed and dense range. For real $z>1$, the Neumann series $(zI-B{)}^{-1}=z^{-1}{\sum }_{n\ge 0}(B\mathrm{/}z{)}^n$ converges in operator norm, so $B-zI$ is invertible as well. Hence $\sigma (B)\subseteq [0,1]$.

By [A1] and [F1], $B$ has its regular spectral PVM $E_B$ on $\sigma (B)$, extended to ambient Borel sets by intersection with $\sigma (B)$ as in the Statement. If $x=E_B(\{0\})y$, then [F4] gives $Bx={\mathrm{\Phi }}_{E_B}(b{\mathbf{1}}_{\{0\}})y=0$, so $E_B(\{0\})H\subseteq \ker B$. Conversely, if $Bx=0$, [F3] gives $0=\mathrm{\parallel }Bx{\mathrm{\parallel }}^2=\int b^{2}d{E}_x(b)$. Within $[0,1]$, $\{b>0\}={\bigcup }_{n\ge 1}[1\mathrm{/}n,1]$ and $b^2\ge n^{-2}$ on each such interval, $E_x([1\mathrm{/}n,1])=0$ for every $n$. Strong countable additivity and $E_B(\sigma (B))=I$ then imply $x=E_B(\{0\})x$. Since $B$ is injective, $E_B(\{0\})=0$. This is a zero spectral projection; it does not remove $0$ from the operator spectrum.

The map $\varphi (b)=b^{-1}-1$ on $\sigma (B)\setminus \{0\}$ is Borel. For a Borel set $S\subseteq [0,\mathrm{\infty })$ set $E_A(S):=E_B(\{b\in \sigma (B)\setminus \{0\}:\varphi (b)\in S\})$. Inverse images preserve disjoint unions and intersections, so [F2] makes $E_A$ a PVM; its total value is $E_B(\sigma (B)\setminus \{0\})=I$ because $E_B(\{0\})=0$. The defining identity on indicator functions, extended first to simple functions and then by uniform approximation, gives ${\mathrm{\Phi }}_{E_A}(f)={\mathrm{\Phi }}_{E_B}(g)$ for every bounded Borel $f$ on $[0,\mathrm{\infty })$, where $g(b)=f(\varphi (b))$ for $b>0$ and $g(0)=0$.

In integrals set $\varphi (0)=0$, and interpret each inverse-power multiplier with a cutoff as zero where its cutoff vanishes. Write ${\mu }_x^B(S)=⟨E_B(S)x,x⟩$. From the definition in step 4.1, ${\mu }_x(S)={\mu }_x^B(\{b>0:\varphi (b)\in S\})$, so $\int {\lambda }^{2}d{\mu }_x(\lambda )=\int \varphi (b{)}^{2}d{\mu }_x^B(b)$. If $x=By$, multiplicativity and the star-homomorphism identity [F4] give $d{\mu }_x^B(b)=b^{2}d{\mu }_y^B(b)$; hence $\int \varphi (b{)}^{2}d{\mu }_x^B(b)={\int }_{(0,1]}(1-b{)}^{2}d{\mu }_y^B(b)\le \mathrm{\parallel }y{\mathrm{\parallel }}^2$. Thus $x$ lies in the domain defining $T_A$. Conversely, if $\int {\varphi }^{2}d{\mu }_x^B<\mathrm{\infty }$, then ${\int }_{(0,1]}{b}^{-2}d{\mu }_x^B=\int (\varphi +1{)}^{2}d{\mu }_x^B\le 2\int {\varphi }^{2}d{\mu }_x^B+2\mathrm{\parallel }x{\mathrm{\parallel }}^2<\mathrm{\infty }$. The bounded truncations $y_N:={\mathrm{\Phi }}_{E_B}(b^{-1}{\mathbf{1}}_{[1\mathrm{/}N,1]})(x)$ are Cauchy by the quadratic identity [F3], since their squared tails are tails of this finite integral, and converge to some $y\in H$. Multiplying by $B$ gives $B{y}_N=E_B([1\mathrm{/}N,1])x\to x$ by strong countable additivity and $E_B(\{0\})=0$, so $By=x$. Hence $D(T_A)=\mathrm{Ran}B=D(A)$. For $x=By$, the truncation in the definition of $T_A$ is, by step 4.1, ${\mathrm{\Phi }}_{E_B}((1-b){\mathbf{1}}_{[1\mathrm{/}(N+1),1]})(y)$. These bounded multipliers converge pointwise off $\{0\}$ to $1-b$ and are uniformly bounded by $1$; [F4] gives strong convergence to ${\mathrm{\Phi }}_{E_B}(1-b)y=(I-B)y=y-x=Ax$. Thus $T_A=A$ on the common domain, proving the operator identity.

If $f$ is bounded Borel on $[0,\mathrm{\infty })$, step 4.1 gives $f(A)={\mathrm{\Phi }}_{E_B}(g)$ for $g(b)=f(\varphi (b))$ on $b>0$ and $g(0)=0$. By [F3], $\mathrm{\parallel }f(A)\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\sup }_{\lambda \ge 0}\mathrm{\mid }f(\lambda )\mathrm{\mid }$, and $\mathrm{\parallel }f(A)u{\mathrm{\parallel }}^2=\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d{\mu }_u$. This proves the bounded Borel calculus claims.

With these parameters, ${\mathrm{\hslash }}^2\mathrm{/}(2m_j)=1$, $m_j{\omega }_j^2\mathrm{/}2=1$, and ${\mathrm{\ell }}_j=\sqrt{\mathrm{\hslash }\mathrm{/}(m_j{\omega }_j)}=1$. The parameter substitution in the inspected proof of [F1] uses precisely this scaling; [F2] specifies the resulting unscaled product basis $h_{\alpha }=g_{{\alpha }_1}(x_1)g_{{\alpha }_2}(x_2)$ and its eigenvalue ${\lambda }_{\alpha }={\sum }_{j=1}^2(2{\alpha }_j+1)=2\mathrm{\mid }\alpha \mathrm{\mid }+2$.

By [F3] each Hermite factor is a polynomial times $e^{-x_j^2\mathrm{/}2}$, so $h_{\alpha }$ and all its derivatives decay rapidly enough for integration by parts against any Schwartz function. The operator $H^r$ is a finite sum of polynomial-coefficient derivatives and maps Schwartz space continuously to itself, hence $H^{r}f\in L^2$. Repeated integration by parts in $x_1$ and $x_2$ has no boundary terms; it gives $⟨H^{r}f,h_{\alpha }⟩=⟨f,H^r{h}_{\alpha }⟩={\lambda }_{\alpha }^r⟨f,h_{\alpha }⟩$.

Parseval for the complete orthonormal basis [F1]–[F2] is supplied by [F4] and gives $\mathrm{\parallel }{H}^{r}f{\mathrm{\parallel }}_2^2={\sum }_{\alpha }\mathrm{\mid }⟨H^{r}f,h_{\alpha }⟩{\mathrm{\mid }}^2={\sum }_{\alpha }(2\mathrm{\mid }\alpha \mathrm{\mid }+2{)}^{2r}\mathrm{\mid }⟨f,h_{\alpha }⟩{\mathrm{\mid }}^2$. Since $1+\mathrm{\mid }\alpha \mathrm{\mid }\le 2\mathrm{\mid }\alpha \mathrm{\mid }+2$, the asserted weighted bound follows.

Each $H^r$ is a finite differential expression with polynomial coefficients, so its Schwartz seminorms bound the pointwise polynomially weighted derivatives in that expression. In two dimensions, $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le C{\sup }_x(1+\mathrm{\mid }x\mathrm{\mid }{)}^2\mathrm{\mid }u(x)\mathrm{\mid }$ because $(1+\mathrm{\mid }x\mathrm{\mid }{)}^{-4}$ is integrable. Thus $f\mapsto \mathrm{\parallel }{H}^{r}f{\mathrm{\parallel }}_2$ is a continuous seminorm on Schwartz space, and the weighted coefficient map is continuous.

Let $I={\mathbb{N}}_0^2$, enumerated by total degree and lexicographic order. For each finite $F\subset I$, let ${\mu }_F$ have density ${\prod }_{\alpha \in F}(2\pi {)}^{-1\mathrm{/}2}{e}^{-x_{\alpha }^2\mathrm{/}2}$ on ${\mathbb{R}}^F$. [F5] makes these probability laws; integrating out coordinates with [F6] proves their consistency. Each coordinate space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is standard Borel by [F4]. Applying [F1] with [A1] yields $\mathrm{\Omega }={\mathbb{R}}^I$ with its cylinder $\sigma$-algebra and a probability $\mathbb{P}$ having these finite marginals. The coordinate variables ${\xi }_{\alpha }(\omega )={\omega }_{\alpha }$ are therefore independent standard normals.

For $p\in {\mathbb{R}}^2$ put $w_m(p)=(\mathrm{\mid }p{\mathrm{\mid }}^2+m^2{)}^{-1\mathrm{/}2}$. Since $m>0$, $w_m$ is smooth, and every derivative is bounded: its derivatives are finite sums of polynomial factors divided by positive powers of $\mathrm{\mid }p{\mathrm{\mid }}^2+m^2$, bounded near the origin and decaying at infinity. The product rule therefore shows multiplication by $w_m$ maps Schwartz space continuously to itself. Define $T_{m}f={\mathcal{F}}_{2\pi }^{-1}(w_m(2\pi \cdot ){\mathcal{F}}_{2\pi }f)$. By [F8], $T_{m}f$ is Schwartz; because $w_m$ is real and even, $T_{m}f$ is real for real $f$. Under the exact convention in the Statement, $\widehat{f}(p)={\mathcal{F}}_{2\pi }f(p\mathrm{/}(2\pi ))$, and the change of variables $p=2\pi \xi$ gives $\widehat{T_{m}f}(p)=w_m(p)\widehat{f}(p)$.

For a finite initial segment $I_N$ of the fixed enumeration set $S_N={\sum }_{\alpha \in I_N}(1+\mathrm{\mid }\alpha \mathrm{\mid }{)}^{-4}{\xi }_{\alpha }^2$. The density in [F5] is normalized, and integration by parts using $(-x{e}^{-x^2\mathrm{/}2}{)}^{\mathrm{\prime }}=(x^2-1)e^{-x^2\mathrm{/}2}$ gives $\mathbb{E}{\xi }_{\alpha }^2=1$. Hence $\mathbb{E}{S}_N={\sum }_{\alpha \in I_N}(1+\mathrm{\mid }\alpha \mathrm{\mid }{)}^{-4}\le {\sum }_{n\ge 0}(n+1)(n+1{)}^{-4}={\sum }_{n\ge 0}(n+1{)}^{-3}=:C<\mathrm{\infty }$: grouping $n+1$ in dyadic blocks $[2^j,2^{j+1})$ bounds each block by $2^j{2}^{-3j}=2^{-2j}$, a summable geometric series. Markov's inequality, which follows from ${\mathbf{1}}_{\{S_N>M\}}\le S_N\mathrm{/}M$, gives $\mathbb{P}(S_N>M)\le C\mathrm{/}M$. Since the events increase to $\{S:={\sum }_{\alpha }(1+\mathrm{\mid }\alpha \mathrm{\mid }{)}^{-4}{\xi }_{\alpha }^2>M\}$, countable additivity gives $\mathbb{P}(S>M)\le C\mathrm{/}M$; letting integer $M\to \mathrm{\infty }$ shows $S<\mathrm{\infty }$ on a common event ${\mathrm{\Omega }}_0$ of probability one.

For $f\in {\mathcal{S}}_{\mathbb{R}}({\mathbb{R}}^2)$ put $c_{\alpha }(f)=⟨f,h_{\alpha }⟩$. On ${\mathrm{\Omega }}_0$, [F11] and [F3] give ${\sum }_{\alpha }\mathrm{\mid }{\xi }_{\alpha }{c}_{\alpha }(f)\mathrm{\mid }\le S^{1\mathrm{/}2}({\sum }_{\alpha }(1+\mathrm{\mid }\alpha \mathrm{\mid }{)}^4\mathrm{\mid }{c}_{\alpha }(f){\mathrm{\mid }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Thus $W_0(f):={\sum }_{\alpha }{\xi }_{\alpha }{c}_{\alpha }(f)$ converges absolutely for every $f$ on the same event, is real-linear, and satisfies $\mathrm{\mid }{W}_0(f)\mathrm{\mid }\le S^{1\mathrm{/}2}\mathrm{\parallel }{H}^{2}f{\mathrm{\parallel }}_2$. The last quantity is a continuous Schwartz seminorm by [F3], so each sample is a tempered distribution. Set $W_0=0$ outside ${\mathrm{\Omega }}_0$. Every evaluation $W_0(f)$ is measurable as the almost-sure limit of measurable finite sums, hence $W_0$ is measurable for the cylinder $\sigma$-algebra.

The partial sums for $W_0(f)$ converge in $L^2(\mathrm{\Omega },\mathbb{P})$ as well: independence, unit variance, and Parseval in [F10] make their squared $L^2$ tail equal to the corresponding tail of ${\sum }_{\alpha }\mathrm{\mid }{c}_{\alpha }(f){\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. For any finite real combination ${\sum }_j{t}_j{W}_0(f_j)$, a finite partial sum is a linear combination of independent standard normals, with characteristic function ${\prod }_{\alpha }{e}^{-t^2{c}_{\alpha }^2\mathrm{/}2}=e^{-t^2{\sum }_{\alpha }{c}_{\alpha }^2\mathrm{/}2}$; by [F10] its variance converges to $\mathrm{\parallel }{\sum }_j{t}_j{f}_j{\mathrm{\parallel }}_2^2$. If $X_N\to X$ in $L^2$, then $\mathrm{\mid }\mathbb{E}{e}^{it{X}_N}-\mathbb{E}{e}^{itX}\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }\mathbb{E}\mathrm{\mid }{X}_N-X\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }\mathrm{\parallel }{X}_N-X{\mathrm{\parallel }}_2\to 0$, so the limit has the centered normal characteristic function with that variance. For positive variance this is the characteristic function of the density defining $N(0,v)$, by [F5] and scaling; at zero variance it is the characteristic function of the point mass at zero. Applying [F13] to these Borel probability laws identifies the limit law. Hence the family is jointly Gaussian, centered, and $\mathbb{E}[W_0(f)W_0(g)]=\int f(x)g(x)d^{2}x$.

Set $\mathrm{\Phi }(f)=W_0(T_{m}f)$. Each sample in ${\mathrm{\Omega }}_0$ is a continuous real-linear functional of $f$ because both $W_0$ and $T_m$ are continuous on Schwartz space; outside ${\mathrm{\Omega }}_0$ it is zero. Thus $\mathrm{\Phi }$ is a measurable ${\mathcal{S}}_{\mathbb{R}}^{\mathrm{\prime }}$-valued random variable for the cylinder $\sigma$-algebra. From step 4.1 its covariance is $\mathbb{E}[\mathrm{\Phi }(f)\mathrm{\Phi }(g)]=\int T_{m}f(x)T_{m}g(x)d^{2}x$. Applying [F9] and changing variables $p=2\pi \xi$ gives $(2\pi {)}^{-2}\int \widehat{T_{m}f}(p)\overline{\widehat{T_{m}g}(p)}{d}^{2}p$. Since $f,g$ are real, $\widehat{f}(-p)=\overline{\widehat{f}(p)}$ and $\widehat{g}(-p)=\overline{\widehat{g}(p)}$; the substitution $p\mapsto -p$ changes this to $(2\pi {)}^{-2}\int \overline{\widehat{f}(p)}\widehat{g}(p)w_m(p{)}^2{d}^{2}p=C_m(f,g)$ by step 1.2. The finite-dimensional distributions remain centered Gaussian by step 4.1 and linearity of $T_m$.

In ${\mathfrak{l}}_{\mathbb{C}}$ set $L_i=(J_i+i{K}_i)\mathrm{/}2$ and $R_i=(J_i-i{K}_i)\mathrm{/}2$. Expanding the stated brackets gives $[L_i,L_j]={ϵ}_{ijk}{L}_k$, $[R_i,R_j]={ϵ}_{ijk}{R}_k$, and $[L_i,R_j]=0$. The real span $\mathfrak{k}={\mathrm{span}}_{\mathbb{R}}\{L_i,R_i\}$ has compact Lie-algebra bracket: with ${\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, ${\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$, ${\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, the basis $-i{\sigma }_i\mathrm{/}2$ of $\mathfrak{s}\mathfrak{u}(2)$ has the same ${ϵ}_{ijk}$ brackets. The two summands commute, and $\mathfrak{k}{\otimes }_{\mathbb{R}}\mathbb{C}={\mathfrak{l}}_{\mathbb{C}}$ because $J_i=L_i+R_i$ and $K_i=(L_i-R_i)\mathrm{/}i$.

Let $G=\mathrm{S}\mathrm{U}(2)\times \mathrm{S}\mathrm{U}(2)$, whose Lie algebra is $\mathfrak{k}$. The matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2=1$ parametrize $\mathrm{S}\mathrm{U}(2)=\{M\in M_2(\mathbb{C}):M^{\ast }M=I,\det M=1\}$ by $(\mathrm{Re}a,\mathrm{Im}a,\mathrm{Re}b,\mathrm{Im}b)\in S^3\subset {\mathbb{R}}^4$, so $G$ is compact. To prove simple connectivity in the form needed here, take any piecewise-smooth based loop in $S^3$. Uniform continuity gives a finite subdivision on which each subarc lies in an open hemisphere. If $\delta (t)$ is the short great-circle arc between its endpoints, then $H(s,t)=((1-s)\gamma (t)+s\delta (t))\mathrm{/}\mathrm{\parallel }(1-s)\gamma (t)+s\delta (t)\mathrm{\parallel }$ is defined throughout that hemisphere, fixes the endpoints, and homotopes the subarc to $\delta$. Each great-circle arc lies in the intersection of $S^3$ with a proper two-dimensional linear subspace $P_j\subset {\mathbb{R}}^4$. A finite union of proper linear subspaces cannot cover ${\mathbb{R}}^4$: induct on their number; after choosing $x$ outside $P_1,\dots ,P_{n-1}$, if $x\mathrm{\notin }{P}_n$ we are done, and otherwise choose $y\mathrm{\notin }{P}_n$; the line $x+ty$ meets each $P_j$ in at most one parameter value, so a real $t$ outside the finite forbidden set gives a point outside all $P_j$. Thus the resulting polygonal loop omits some $q\in S^3$. Stereographic projection identifies $S^3\setminus \{q\}$ with ${\mathbb{R}}^3$; straight-line contraction there to the basepoint fixes the loop endpoints and is piecewise smooth. Hence every piecewise-smooth based loop contracts through piecewise-smooth based loops. Applying this factorwise proves the required property for $G$.